Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-21
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  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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The two-set measure identity μ(A∪B)+μ(A∩B)=μ(A)+μ(B)

Statement

For measurable sets A and B in a measure space,

μ(A∪B)+μ(A∩B)=μ(A)+μ(B).

The equality is valid in [0,+∞], including when one or both sides equal +∞.

Facts & Assumptions

Given: A measure μ and measurable sets A,B.

[L1]

A measure is additive on every finite pairwise disjoint measurable family (Measures on sigma-algebras).

Proof

technique · direct
1.1given

Put C=A∖B, D=A∩B, and F=B∖A. These sets are measurable and pairwise disjoint, with A=C∪D, B=D∪F, and A∪B=C∪D∪F.

2.1step 1.1L1

Finite additivity gives μ(A)=μ(C)+μ(D), μ(B)=μ(D)+μ(F), and μ(A∪B)=μ(C)+μ(D)+μ(F).

3.1step 2.1algebra∎

Adding μ(D)=μ(A∩B) to the last equality and regrouping nonnegative extended sums gives the displayed identity; no subtraction occurs, so infinite values and the cases A=∅, B=∅, or A=B are included.

Depends on

Dependency tree · 2 levels

Sources